This set of Digital Signal Processing Multiple Choice Questions & Answers (MCQs) focuses on “General Consideration for Design of Digital Filters”.

1. The ideal low pass filter cannot be realized in practice.

a) True

b) False

View Answer

Explanation: We know that the ideal low pass filter is non-causal. Hence, a ideal low pass filter cannot be realized in practice.

2. The following diagram represents the unit sample response of which of the following filters?

a) Ideal high pass filter

b) Ideal low pass filter

c) Ideal high pass filter at ω=π/4

d) Ideal low pass filter at ω=π/4

View Answer

Explanation: At n=0, the equation for ideal low pass filter is given as h(n)=ω/π.

From the given figure, h(0)=0.25=>ω=π/4.

Thus the given figure represents the unit sample response of an ideal low pass filter at ω=π/4.

3. If h(n) has finite energy and h(n)=0 for n<0, then which of the following are true?
d) None of the mentioned

View Answer

Explanation: If h(n) has finite energy and h(n)=0 for n<0, then according to the Paley-Wiener theorem, we have

4. If |H(ω)| is square integrable and if the integral is finite, then the filter with the frequency response H(ω)=|H(ω)|e^{jθ(ω)} is?

a) Anti-causal

b) Constant

c) Causal

d) None of the mentioned

View Answer

Explanation: If |H(ω)| is square integrable and if the integral is finite, then we can associate with |H(ω)| and a phase response θ(ω), so that the resulting filter with the frequency response H(ω)=|H(ω)|e

^{jθ(ω)}is causal.

a) True

b) False

View Answer

Explanation: One important conclusion that we made from the Paley-Wiener theorem is that the magnitude function |H(ω)| can be zero at some frequencies, but it cannot be zero over any finite band of frequencies, since the integral then becomes infinite. Consequently, any ideal filter is non-causal.

6. If h(n) is causal and h(n)=he(n)+ho(n),then what is the expression for h(n) in terms of only h_{e}(n)?

a) h(n)=2h_{e}(n)u(n)+h_{e}(0)δ(n) ,n ≥ 0

b) h(n)=2h_{e}(n)u(n)+h_{e}(0)δ(n) ,n ≥ 1

c) h(n)=2h_{e}(n)u(n)-h_{e}(0)δ(n) ,n ≥ 1

d) h(n)=2h_{e}(n)u(n)-h_{e}(0)δ(n) ,n ≥ 0

View Answer

Explanation: Given h(n) is causal and h(n)= h

_{e}(n)+h

_{o}(n)

=>h

_{e}(n)=1/2[h(n)+h(-n)] Now, if h(n) is causal, it is possible to recover h(n) from its even part he(n) for 0≤n≤∞ or from its odd component h

_{o}(n) for 1≤n≤∞.

=>h(n)= 2h

_{e}(n)u(n)-h

_{e}(0)δ(n) ,n ≥ 0

7. If h(n) is causal and h(n)=he(n)+ho(n),then what is the expression for h(n) in terms of only h_{o}(n)?

a) h(n)=2h_{o}(n)u(n)+h(0)δ(n) ,n ≥ 0

b) h(n)=2h_{o}(n)u(n)+h(0)δ(n) ,n ≥ 1

c) h(n)=2h_{o}(n)u(n)-h(0)δ(n) ,n ≥ 1

d) h(n)=2h_{o}(n)u(n)-h(0)δ(n) ,n ≥ 0

View Answer

Explanation: Given h(n) is causal and h(n)= h

_{e}(n)+h

_{o}(n)

=>h

_{e}(n)=1/2[h(n)+h(-n)] Now, if h(n) is causal, it is possible to recover h(n) from its even part h

_{e}(n) for 0≤n≤∞ or from its odd component h

_{o}(n) for 1≤n≤∞.

=>h(n)= 2h

_{o}(n)u(n)+h(0)δ(n) ,n ≥ 1

since ho(n)=0 for n=0, we cannot recover h(0) from h

_{o}(n) and hence we must also know h(0).

8. If h(n) is absolutely summable, i.e., BIBO stable, then the equation for the frequency response H(ω) is given as?

a) H_{I}(ω)-j H_{R}(ω)

b) H_{R}(ω)-j H_{I}(ω)

c) H_{R}(ω)+j H_{I}(ω)

d) H_{I}(ω)+j H_{R}(ω)

View Answer

Explanation: . If h(n) is absolutely summable, i.e., BIBO stable, then the frequency response H(ω) exists and

H(ω)= H

_{R}(ω)+j H

_{I}(ω)

where H

_{R}(ω) and H

_{I}(ω) are the Fourier transforms of he(n) and ho(n) respectively.

9. H_{R}(ω) and H_{I}(ω) are interdependent and cannot be specified independently when the system is causal.

a) True

b) False

View Answer

Explanation: Since h(n) is completely specified by he(n), it follows that H(ω) is completely determined if we know HR(ω). Alternatively, H(ω) is completely determined from HI(ω) and h(0). In short, H

_{R}(ω) and H

_{I}(ω) are interdependent and cannot be specified independently when the system is causal.

10. What is the Fourier transform of the unit step function U(ω)?

a) πδ(ω)-0.5-j0.5cot(ω/2)

b) πδ(ω)-0.5+j0.5cot(ω/2)

c) πδ(ω)+0.5+j0.5cot(ω/2)

d) πδ(ω)+0.5-j0.5cot(ω/2)

View Answer

Explanation: Since the unit step function is not absolutely summable, it has a Fourier transform which is given by the equation

U(ω)= πδ(ω)+0.5-j0.5cot(ω/2)

_{I}(ω) is uniquely determined from H

_{R}(ω) through the integral relationship. This integral is called as Continuous Hilbert transform.

a) True

b) False

View Answer

Explanation: If the H

_{I}(ω) is uniquely determined from HR(ω) through the integral relationship. This integral is called as discrete Hilbert transform.

12. The magnitude |H(ω)| cannot be constant in any finite range of frequencies and the transition from pass-band to stop-band cannot be infinitely sharp.

a) True

b) False

View Answer

Explanation: Causality has very important implications in the design of frequency-selective filters. One among them is the magnitude |H(ω)| cannot be constant in any finite range of frequencies and the transition from pass-band to stop-band cannot be infinitely sharp. This is a consequence of Gibbs phenomenon, which results from the truncation of h(n) to achieve causality.

13. The frequency ω_{P} is called as:

a) Pass band ripple

b) Stop band ripple

c) Pass band edge ripple

d) Stop band edge ripple

View Answer

Explanation: Pass band edge ripple is the frequency at which the pass band starts transiting to the stop band.

14. Which of the following represents the bandwidth of the filter?

a) ω_{P}+ ω_{S}

b) -ω_{P}+ ω_{S}

c) ω_{P}-ω_{S}

d) None of the mentioned

View Answer

Explanation: If ωP and ωS represents the pass band edge ripple and stop band edge ripple, then the transition width -ω

_{P}+ ω

_{S}

_{S}gives the bandwidth of the filter.

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